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I am quite new to analysis and I have a basic question. Whenever the Riemann integral is defined, the Lebesgue integral is defined and the two are the same. Perhaps the Riemann integral is easier to construct. Besides this, are there any theoretical reasons to consider the Riemann Integral instead of the Lebesgue?

For instance, what is an important property that the Riemann integrable functions have?

I'm also wondering why it is the Riemann integral that is predominantly taught and not the Lebesgue.

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    $\begingroup$ Riemann integral is simpler to construct than Lebesgue integral, this is the main reason to teach Riemann first. Also Riemann integration can be extended to improper integrals of Riemann, and many of them are not defined under the Lebesgue theory $\endgroup$ – Masacroso Apr 26 at 0:15
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    $\begingroup$ It is easier to teach Riemann integral to students with little background in concepts like compact metric spaces etc. Measure theory requires some amount of Mathematical maturity but it surely makes Analysis a lot easier. I use measure theory everyday but I use Riemann integration very very rarely. $\endgroup$ – Kabo Murphy Apr 26 at 0:15
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    $\begingroup$ See my answer to another question ... math.stackexchange.com/a/3053942/442 $\endgroup$ – GEdgar Apr 26 at 0:27
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    $\begingroup$ Riemann integration is generally pretty terrible, but it doesn't require much beyond basic real analysis to define. Like @KaviRamaMurthy, I use measure theory daily but have no occasion to use Riemann integration. $\endgroup$ – anomaly Apr 26 at 0:40
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    $\begingroup$ In numerical integration, practically all quadrature rules are developed from the perspective of Riemann integration. There are some Monte Carlo or quasi Monte Carlo algorithms for Lebesgue integration. I am not familiar with them, but my impression is that they are slow. $\endgroup$ – user1551 Apr 26 at 1:07
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You asked for an important property of Riemann integrable functions.

A bounded function defined on $[a,b]$ is Riemann integrable iff the set of its points of discontinuity has measure zero.

Note functions like $\mathbb{1}_\mathbb{Q}$ do not meet this criterion (for nontrivial intervals), but are still Lebesgue integrable. The criterion for a bounded function on $[a,b]$ to be Lebesgue integrable is that the function be measurable, which requires defining a measure, which takes you on a medium-sized detour.*

Lebesgue integration applies to strictly more functions than Riemann does**. And all the same limiting processes are possible: two-sided improper, or requiring that the limit from either side exist individually, including when the point is $\infty$.

Why use the Riemann integral at all? It's a good example of a mathematical construction through a limiting process, which allows teachers to open the black box of the integral without undue pain. It arguably is the more intuitive definition.


*-"[Lebesgue] measure zero" is easier to explain than an entire measure.

**-There is an exception which is an artifact of the definitions. For an integral like $$\int_0^\infty \frac{\sin x}{x}$$ the convention for Riemann integration would be to define the limit as $$\lim_{t \to \infty} \int_0^t \frac{\sin x}{x}.$$ If we accept this definition of the limit, the limit exists for Riemann or for Lebesgue. In the Lebesgue integral, it is more usual to place a more stringent requirement on the function, namely that $\int_0^\infty \max(f, 0)$ and $\int_0^\infty -\min(f, 0)$ exist separately. In this sense both Riemann and Lebesgue are divergent. But because the Lebesgue integral mostly uses the second definition, one can say somewhat misleadingly that "the improper Riemann integral exists but not the improper Lebesgue integral".

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    $\begingroup$ +1, maybe you can add that there are functions as $\frac{\sin x}{x}$ which are improper Riemann integrable but not improper Lebesgue integrable. $\endgroup$ – EuklidAlexandria Apr 26 at 14:12
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    $\begingroup$ One reason for choosing the second definition is that all Lebesgue integrable functions will then form a vector space. Otherwise it cannot be guaranteed that $-f$ is also integrable if $f$ is. $\endgroup$ – EuklidAlexandria Apr 27 at 11:06

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